Question: Evaluate the double integral. $ \int_0^2 \int_{-x}^{2x} 4x + 2y \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $56$ (Choice B) B $64$ (Choice C) C $40$ (Choice D) D $24$
Explanation: First, we evaluate the inner integral. We can substitute in the $-x$ and $2x$ at the end as if they were numerical bounds. $\begin{aligned} \int_0^2 \int_{-x}^{2x} 4x + 2y \, dy \, dx &= \int_0^2 \left[ 4xy + y^2 \right]_{-x}^{2x} \, dx \\ \\ &= \int_0^2 8x^2 + 4x^2 - (-4x^2 + x^2) \, dx \\ \\ &= \int_0^2 15x^2 \, dx \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^2 15x^2 \, dx &= \left[ 5x^3 \right])0^2 \\ \\ &= 40 \end{aligned}$ The answer: $ \int_0^2 \int_{-x}^{2x} 4x + 2y \, dy \, dx = 40$